I have recently acquired a couple of mysterious ultra/super capacitors from my brother. Apparently he doesn't remember any of the specifications or even brand... To further complicate matters, they have no meaningful identification information stamped or printed on them. (There is a bar code label with alphanumeric code but a quick Google search using it found nothing.)

Looks like it's time to fire up the Scooby-Doo Mystery Buss, 'cause were going on an adventure folks.

First, I figured I'd try to measure the capacitance. Since my LCR meter isn't specified for enormous capacitors like these, I had to get creative with my test equipment.

Taking basic physics into account, we have that capacitance is proportional to the stored charge per volt across the capacitor:

$$ C=\frac{q}{V} $$

where the accumulated charge in the capacitor is the integral of the current through the capacitor:

$$ \int i(t)dt=q$$

Using a current source to charge the capacitor we can simplify the calculations, using only delta measurements of the charge and voltage across the capacitor.

$$ C=\frac{\Delta q}{\Delta V}=\frac{i\Delta t}{\Delta V} $$

With my Advantest R6144 current source I can then charge the capacitor at a set current and simply measure the voltage across the capacitor using my Tektronix DMM4050 in the trendplot mode.

Picures of Test Setup

However, this is where I start to see some rather large numbers. It's possible the capacitor really is ~2200 farads, but that seems a bit high. Admittedly, the capacitor is quite large at ~5.5" long by ~1" radius.

And now some questions for the fine folks of Electrical Engineering Stack Exchange: Is this method a viable means to measure super capacitors? Or is there a more suitable method that I can apply to measure them? Also, does the capacitance of super/ultra capacitors significantly change vs. voltage of the capacitor? E.g., are these measured results predictive/indicative for higher charge voltages. I would reckon the capacitance should fluctuate some, but I doubt its that much. Probably at worst it's a few hundred farads, but I'm no expert on the matter.

Also, and somewhat more importantly, how would I find the maximum charge voltage without destroying the capacitor? Would a constant current charge of say 100uA over a few weeks till the voltage reaches some sort of equilibrium with self discharge work. Then back off a couple hundred milivolts and call that the max charge voltage. Or will it just reach a tripping point and self-destruct while spraying electrolyte all over my lab?

Finally, how do you determine the polarity orientation of the capacitors? These are not marked in any way, and both terminals are identical. I cast my bet with the residual voltage stored in the capacitor. I assume the dielectric absorption/memory effect from previous charging knows the correct direction...

At any rate, it's sort of fun to try and determine the characteristics of these capacitors. But it's still a touch aggravating that there are no useful markings on them, like polarity orientation, manufacturer, ect.

  • \$\begingroup\$ Looking at the pdfs that Dan1138 kindly provided, I believe that a constant current charge of 1mA to 100uA (after the cap has been charged to ~2.5V under a much faster rate) could indeed ballpark the maximum charging voltage. If the leakage current at rated voltage is close to 4.2mA (for Maxwell 2000F super cap.), then a constant current of any value less than that should never over charge the capacitor since the leakage does not charge the capacitor. Let me know what you guys think. \$\endgroup\$
    – Big Gulps
    Jul 23, 2018 at 5:21
  • \$\begingroup\$ 2200F is the correct order of magnitude for an ultracapacitor. Also they all seem to have the same maximum voltage. \$\endgroup\$
    – user253751
    Jul 23, 2018 at 6:54
  • \$\begingroup\$ Can you edit your question and inline your image please? For those of us who live behind a proxy we can't see it. \$\endgroup\$
    – UKMonkey
    Jul 23, 2018 at 11:47
  • \$\begingroup\$ Leakage current may upset the measurements on charging, but if a discharge measurement also says 2200 uF then it's probably true. \$\endgroup\$ Jul 23, 2018 at 16:22
  • \$\begingroup\$ After responding to user194292, I realized you could automate the capacitance measurement by the fact that $$V(t)=\frac{i_{cc}}{C}t+V_0$$, which is a linear equation of the form Mx+b. Using a daq system you need only measure V(t) and then can compute capacitance as $$C=i_{cc}M^{-1}$$ from the slope of the data (obtained from linear regression). And alternately, you can use a DMM in the ohms mode as talked about below. However, the accepted method is to discharge the cap from Vmax to 50% Vmax under CC to obtain the working capacitance, rather than small signal capacitance. \$\endgroup\$
    – Big Gulps
    Jul 24, 2018 at 0:24

3 Answers 3


This is Maxwell's process for measuring C from their test spec.

enter image description here

\$C=C_{dcd}=\dfrac{I_5*(t_5-t_4)}{V_5-V_4}\$ The Capacitance is charged and discharged at rated current but measured by the rated from Urated to 50% Urated.

Note that the voltage sags towards the previous voltage due to an additional RC time constant in parallel. (i.e. memory effect) Here it is show to be around 5% of full scale of 10% for a half voltage discharge. This memory effect indicates another "double layer electric effect" capacitance between 5% and 10% of C.

What this means is like in batteries, if you charge and discharge much slower ( at least 10x slower) then the storage capacity increases 5~10%, similar to the best low ESR Li-ion batteries , which are advertised as having no memory effects ( relative to NiCad.)

  • \$\begingroup\$ It's interesting to see that they use a double charge/discharge cycle and perform the actual measurement on the final discharge. I think for most people this isn't a practical method of measurement--who has a test fixture that can output 100 A constant current pulses and the daq system to capture it all. That said, I think I'll throw a couple mosfets in parallel with some 100mohm resistors and opamp for CC and use my oscilloscope to capture the delta measurement for the discharge cycle. Regardless, I think my low current method works for ballpark measurements. \$\endgroup\$
    – Big Gulps
    Jul 23, 2018 at 4:35
  • 3
    \$\begingroup\$ well, if I was to test supercaps for current delivery properties, then well, I might need a beefy power supply; in that sense "who has a test fixture like that" is "people who really need to measure systems under high current, probably, including folks who test supercaps". \$\endgroup\$ Jul 23, 2018 at 4:56
  • 1
    \$\begingroup\$ A large battery can provide the current with a good MOSFET, but a low current test may be 10% higher C due to the secondary capacitance supporting the output with a much lower dV/dt. \$\endgroup\$ Jul 23, 2018 at 5:00

From the images of the cell in test setup photo they appear to be similar to the Maxwell DuraBlue line of Ultracapacitors. See this datasheet for more information.

The Maxwell application note 1007239, Test Procedures for Capacitance, ESR, Leakage Current and Self-Discharge Characterizations of Ultracapacitors, may be helpful.

This line of "supercapacitors" has a maximum working voltage of 2.85 VDC and a typical capacitance of 3400 Farads. Most other "supercapacitors" in this type of package have a maximum working voltage of 2.7 VDC.

Be cautious an internal short in these devices can result in a spectacular failure event. You may want to have non-conductive, non-water based fire suppression system available (sand, chemical, CO2, Halon etc.).

Based on the posted test setup photos you will likely melt the alligator clips before you exceed the maximum charge or discharge current.


My usual way is to measure the resistance with a normal multimeter. Assuming that the test voltage/current is more or less continuously applied, you'll see the "resistance" reading increase in comparatively linear manner over time. Rougly averaging this increase in the unit "Ohm per second" gives you the inverse of the capacity.

For example, if the readout increases about 10 Ohms every second, the capacity is about 0.1F. You should check with some known capacities first that your multimeter is of the continuously measuring type where this approximation is good enough.

  • \$\begingroup\$ Most multimeters use a CC source to measure ohms. So this is essentially the same method I've used and outlined above. Measured resistance at DMM is $$R(t)=\frac{V(t)}{i_{cc}}$$, where $$V(t)=\frac{i_{cc}}{C}t + V_0$$ Simplifying, and using delta measurements, gives $$\Delta R(t)=\frac{\Delta t}{C}$$ which implies $$C=\frac{\Delta t}{\Delta R(t)}$$.The only problem I see, is that you don't have time stamps for delta t. Unless the meter has a trendplot (or similar) you'll only get a very rough approximation. But, like you say, this is an easiy way to check without needing expensive equipment. \$\endgroup\$
    – Big Gulps
    Jul 23, 2018 at 23:46

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